A beam of electromagnetic radiation of intensity 6.4 × 10 –5 W/cm² is comprised of wavelength, λ = 310 nm. It falls normally on a metal (work function φ = 2eV) of surface area of 1 cm². If one in 10³ photons ejects an elctron, total number of electrons ejected in 1 s is 10ˣ. (hc = 1240 eVnm, 1eV = 1.6 × 10 –19 J), then x is _____.

Solution

Physics sir · 4 steps

Power over photon energy counts the photons; then keep only one in a thousand of them.

1. Energy of one photon. E = hc/λ = (1240 eV nm)/(310 nm) = 4 eV = 4 × 1.6 × 10⁻¹⁹ = 6.4 × 10⁻¹⁹ J. This is more than φ = 2 eV, so emission does happen. The work function plays no other part.
2. Power on the plate. Intensity times area gives P = 6.4 × 10⁻⁵ W cm⁻² × 1 cm² = 6.4 × 10⁻⁵ W. That is the energy arriving each second.

   nₚₕₒₜₒₙₛ = P/E = (6.4 × 10⁻⁵)/(6.4 × 10⁻¹⁹) = 10¹⁴ s⁻¹

3. Apply the efficiency. Only one photon in 10³ ejects an electron, so electrons per second = 10¹⁴/10³ = 10¹¹.
4. So. The number of electrons ejected in 1 s is 10¹¹, so x = 11.

The common slip: 14. Forgetting that only one photon in 10³ ejects an electron gives 10¹⁴, so x = 14.

The idea to keep. Photons per second is always power divided by hc/λ. The work function only decides whether emission happens; it does not change how many electrons come out.

Answer from the NTA final answer key. Solution by PrepWiser.